Question

help in trigo..........​

Answer 1

You must be Knowing the following formulas:

★ (1 + tan²A) = sec²A

★ (1 + cot²A) = cosec²A

So, just apply it in the given question.

We get,

→ (tanA/sec²A) + (cotA/cosec²A)

(Now change all "tanA", "cotA", "secA", & "cosecA" in terms of "SinA" & "CosA".
It would be easier.)

→ (sinA/cosA)(cos²A) + (cosA/sinA)(sin²A)
= sinA cosA + cosA sinA
= 2 sinA cosA
= sin 2A

Thankyou!!!

Answer 2

Answer:

[tanA/(1+tan^2A)^2]+[cotA/(1+cot^2A)^2] = sinAcosA

1+tan^2A=sec^2A                       ..(1)

1+cot^2A=cosec^2A                   ..(2)

now, put these value in question

(tanA/sec^4A)+(cotA/cosec^4A)

convert into sin and cos ratios

=sinAcos^3A+cosAsin^3A

=(sinAcosA)(sin^2A=cos^2A)

∴sin^2A+cos^2A=1

=sinAcosA